clearly, d=28 and the sequence is
-44, -16, 12, 40, 68, 96, ...
hint: 71*28 = 1988
but you didn't start at zero.
find the number of terms in the progression, which are less than 2000.
-44, -16, 12, 40, 68, 96, ...
hint: 71*28 = 1988
but you didn't start at zero.
3rd term = ar^2 = 12
6th term = ar^5 = 96
divide them
r^3 = 96/12 = 8
r = 2
then in ar^2 = 12
4a = 12
a = 3
so you have 3 + 6 + 12 + ... +(the last value before 2000)
so term(n) = ar^(n-1) < 2000
3(2^(n-1) ) = ..
if n = 9, 3(2^8) = 768
n = 10 , 3(2^9) = 1536
n = 11 , 3(2^10) = 3072
so there are 10 terms
Sum(10) = 3 (2^10 - 1)/(2-1) = 3069
Let's denote the first term of the geometric progression as 'a' and the common ratio as 'r'. Based on the given information, we can write the following equations:
a * r^2 = 12 (Third term)
a * r^5 = 96 (Sixth term)
To find the common ratio, we can divide the second equation by the first equation:
(a * r^5) / (a * r^2) = 96 / 12
r^3 = 8
r = 2
Now that we have the common ratio (r = 2), we can determine the terms of the progression that are less than 2000.
The nth term of the geometric progression can be calculated using the formula:
Tn = a * r^(n-1)
We need to find the highest value of n for which Tn is less than 2000. Let's substitute 2000 in the formula and solve for n:
2000 = a * 2^(n-1)
To simplify the equation, we can divide both sides by 'a':
2000/a = 2^(n-1)
Since 2 is a power of 2, we can rewrite the equation as:
1000/a = 2^n
Take the logarithm of both sides:
log(1000/a) = n * log(2)
Solve for n:
n = log(1000/a) / log(2)
We know that the common ratio (r) is 2, so the third term of the progression is given by:
12 = a * 2^2
12 = 4a
a = 3
Substituting the value of a in the equation for n:
n = log(1000/3) / log(2)
Using a calculator, we can determine the value of n approximately.
Hence, the number of terms in the geometric progression that are less than 2000 is approximately the calculated value of n.